Higher Tier - Algebra revision

IndicesExpanding single bracketsExpanding double bracketsSubstitutionSolving equationsSolving equations from angle probsFinding nth term of a sequenceSimultaneous equations – 2 linearSimultaneous equations – 1 of eachInequalitiesFactorising – common factorsFactorising – quadraticsFactorising – grouping & DOTS

Solving quadratic equations Using the formulaCompleting the squareRearranging formulaeAlgebraic fractions Curved graphsGraphs of y = mx + cGraphing inequalitiesGraphing simultaneous equations Graphical solutions to equationsExpressing laws in symbolic formGraphs of related functionsKinematics

Contents:

Indices

(F2)4

t2 t2

b1

a2 x a3 c0 a4

x7 x4

2e7 x

3ef2

4xy3 2xy

5p5qr x 6p2q6r

Expanding single brackets

e.g. 4(2a + 3) =

x

8a

x

+ 12

Remember to multiply all the terms inside the bracket by the term

immediately in front of the bracket

If there is no term in front of the bracket, multiply by 1 or -1

Expand these brackets and simplify wherever possible: 1. 3(a - 4) =2. 6(2c + 5) =3. -2(d + g) =4. c(d + 4) =5. -5(2a - 3) =6. a(a - 6) =

3a - 1212c + 30-2d - 2gcd + 4c-10a + 15

a2 - 6a

7. 4r(2r + 3) =8. - (4a + 2) =9. 8 - 2(t + 5) =10. 2(2a + 4) + 4(3a + 6) =

11. 2p(3p + 2) - 5(2p - 1) =

8r2 + 12r-4a - 2

-2t - 2

16a + 32

6p2 - 6p + 5

Expanding double brackets Split the double brackets into 2 single brackets and then expand

each bracket and simplify

(3a + 4)(2a – 5)

= 3a(2a – 5) + 4(2a – 5)

= 6a2 – 15a + 8a – 20

“3a lots of 2a – 5 and 4 lots of 2a – 5”

= 6a2 – 7a – 20 If a single bracket is squared (a + 5)2 change it into double

brackets (a + 5)(a + 5)

Expand these brackets and simplify : 1. (c + 2)(c + 6) =2. (2a + 1)(3a – 4) =3. (3a – 4)(5a + 7) =4. (p + 2)(7p – 3) =

c2 + 8c + 126a2 – 5a – 415a2 + a – 28

7p2 + 11p – 6

5. (c + 7)2 =6. (4g – 1)2 =

c2 + 14c + 4916g2 – 8g + 1

Substitution If a = 5 , b = 6 and c = 2 find the value of :

3a ac

4bca

(3a)2

a2 –3b

c2

(5b3 – ac)2

ab – 2c c(b – a)

4b2

Now find the value of each of these expressions if a = - 8 , b = 3.7 and c = 2/3

15 4 144 10

26 2 225

7 9.6 1 144 900

Solving equationsSolve the following equation to find the value of x : 4x + 17 = 7x – 1

17 = 7x – 4x – 117 = 3x – 1

17 + 1 = 3x18 = 3x18 = x 3

6 = xx = 6

Take 4x from both sides

Add 1 to both sides

Divide both sides by 3Now solve these: 1. 2x + 5 = 172. 5 – x = 23. 3x + 7 = x + 154. 4(x + 3) = 20 5

Some equations cannot be solved in this way and “Trial and Improvement” methods are required

Find x to 1 d.p. if: x2 + 3x = 200 Try Calculation Comment

x = 10x = 13

(10 x 10)+(3 x 10) = 130 Too lowToo high(13 x 13)+(3 x 13) = 208

etc.

Solving equations from angle problems

Rule involved:Angles in a quad = 3600

4y + 2y + y + 150 = 360 7y + 150 = 360

7y = 360 – 150 7y = 210 y = 210/7 y = 300

4y

1500y

2y

Find the size of each angle

Angles are:300,600,1200,1500

2v + 394v + 5

Find the value of v 4v + 5 = 2v + 39

4v - 2v + 5 = 39 2v + 5 = 39

2v = 39 - 5 2v = 34

v = 34/2 v = 170

Check: (4 x 17) + 5 = 73 , (2 x 17) + 39 = 73

Rule involved:“Z” angles are

equal

Finding nth term of a simple sequence

This sequence is the 2 times table

shifted a little 5 , 7 , 9 , 11 , 13 , 15 ,…….……

Position number (n)

1 2 3 4 5 6

Each term is found by the position number times 2 then add another 3. So the rule for the sequence is nth term = 2n + 3

2 4 6 8 10 12

Find the rules of these sequences

1, 3, 5, 7, 9,… 6, 8, 10, 12,

……. 3, 8, 13, 18,…… 20,26,32,38,

……… 7, 14, 21,28,

……

1, 4, 9, 16, 25,…

3, 6,11,18,27…….

20, 18, 16, 14,…

40,37,34,31,………

6, 26,46,66,……

100th term = 2 x 100 + 3 = 203

2n – 12n + 45n – 26n + 147n

And these sequences

n2

n2 + 2-2n + 22-3n + 4320n - 14

Finding nth term of a more complex sequence

4 , 13 , 26 , 43 , 64 ,…….……

+9 +13 +17 +21+4 +4 +4

2nd difference is 4 means that the first term is 2n2

1 2 3 4 5n =

2n2 = 2 , 8 , 18 , 32 , 50 ,…….……

Wha

t’s

left = 2 , 5 , 8 , 11 , 14 ,…….……

This sequence has a rule

= 3n - 1

So the nth term = 2n2 + 3n - 1

Find the rule for these

sequences

(a) 10, 23, 44, 73, 110, …(b) 0, 17, 44, 81, 128, …(c) 3, 7, 17, 33, 55, …

(a) nth term = 4n2 + n + 5(b) nth term = 5n2 + 2n – 7 (c) nth term = 3n2 – 5n + 5

Simultaneous equations – 2 linear equations

4a + 3b = 176a 2b = 6

1 Multiply the equations up until the second unknowns have the same sized number in front of them

x 2

x 3

8a + 6b = 3418a 6b = 18

2 Eliminate the second unknown by combining the 2 equations using either SSS or SDA

+26a = 52

a = 52 26 a = 2

3 Find the second unknown by substituting back into one of the equations Put a = 2 into: 4a + 3b = 17

8 + 3b = 173b = 17 - 83b = 9b = 3

So the solutions are:a = 2 and b = 3

Now solve: 5p + 4q = 242p + 5q = 13

Simultaneous equations – 1 linear and 1 quadratic

Sometimes it is better to use a substitution method rather than the elimination method described on the previous slide.

Follow this method closely to solve this pair of simultaneous equations: x2 + y2 = 25 and x + y = 7

Step 1 Rearrange the linear equation: x = 7 - y

Step 2 Substitute this into the quadratic: (7 - y)2 + y2 = 25

Step 3 Expand brackets, rearrange, (7 - y)(7 - y) + y2 = 25 factorise and solve: 49 - 14y + y2 + y2 = 25

2y2 - 14y + 49 = 25 2y2 - 14y + 24 = 0 (2y - 6)(y - 4) = 0 y = 3 or y = 4

Step 4 Substitute back in to find other unknown: y = 3 in x + y = 7 x = 4 y = 4 in x + y = 7 x = 3

InequalitiesInequalities can be solved in exactly the same

way as equations

14 2x – 814 + 8 2x

22 2x

11 x

22 x 2

x 11

Add 8 to both sides

Divide both sides by 2

Remember to turn the sign round as well

The difference is that inequalities can be given

as a range of results

Or on a scale:

Here x can be equal to : 11, 12, 13, 14, 15, ……

8 9 10 11 12 13 14

1. 3x + 1 > 42. 5x – 3 123. 4x + 7 < x + 134. -6 2x + 2 < 10

Find the range of solutions for these inequalities :

X > 1 X = 2, 3, 4, 5, 6 ……orX 3 X = 3, 2, 1, 0, -1 ……orX < 2 X = 1, 0, -1, -2, ……or-4 X < 4 X = -4, -3, -2, -1, 0, 1, 2, 3or

Factorising – common factorsFactorising is basically the

reverse of expanding brackets. Instead of removing brackets you are putting them in and placing all the common

factors in front.

5x2 + 10xy = 5x(x + 2y)

Factorising

Expanding

Factorise the following (and check by expanding):

15 – 3x = 2a + 10 = ab – 5a = a2 + 6a = 8x2 – 4x =

3(5 – x)2(a + 5)a(b – 5)a(a + 6)4x(2x – 1)

10pq + 2p = 20xy – 16x = 24ab + 16a2 = r2 + 2 r = 3a2 – 9a3 =

2p(5q + 1)4x(5y - 4)8a(3b + 2a)r(r + 2)3a2(1 – 3a)

Factorising – quadratics Here the factorising is the reverse of expanding double brackets

x2 + 4x – 21 = (x + 7)(x – 3)

Factorising

Expanding

Factorise x2 – 9x - 22

To help use a 2 x 2 box

x2

- 22

xx

Factor pairs

of - 22:-1, 22- 22, 1- 2, 11- 11, 2Find the

pair which add to give - 9

x2

-22

xx

-11-11x

2x2

Answer = (x + 2)(x – 11)

Factorise the following:

x2 + 4x + 3 = x2 - 3x + 2 = x2 + 7x - 30 = x2 - 4x - 12 = x2 + 7x + 10 =

(x + 3)(x + 1)(x – 2)(x – 1) (x + 10)(x – 3)(x + 2)(x – 6) (x + 2)(x + 5)

Factorising - quadratics When quadratics are more difficult to factorise use this method

Factorise 2x2 + 5x – 3 Write out the factor pairs of – 6 (from 2 multiplied – 3)

-1, 6-6, 1-2, 3-3, 2

Find the pair which add to give + 5

(-1, 6)

Rewrite as 2x2 – 1x + 6x – 3

Factorise x(2x – 1) + 3(2x – 1) in 2 parts Rewrite as (x + 3)(2x – 1)doublebrackets

Now factorise these:(a) 25t2 – 20t + 4(b) 4y2 + 12y + 5(c) g2 – g – 20(d) 6x2 + 11x – 10(e) 8t4 – 2t2 – 1

Answers: (a) (5t – 2)(5t – 2) (b) (2y + 1)(2y + 5)(c) (g – 5)(g + 4) (d) (3x – 2)(2x + 5) (e) (4t2 + 1)(2t2 – 1)

Factorising – grouping and difference of two squares

Fully factorise this expression:6ab + 9ad – 2bc – 3cd

Factorise in 2 parts 3a(2b + 3d) – c(2b + 3d)

Rewrite as double brackets (3a – c)(2b + 3d)

Fully factorise these:(a) wx + xz + wy + yz(b) 2wx – 2xz – wy + yz(c) 8fh – 20fi + 6gh – 15gi

Answers:(a) (x + y)(w + z)(b) (2x – y)(w – z)(c) (4f + 3g)(2h – 5i)

Grouping into pairs

Fully factorise this expression:4x2 – 25

Look for 2 square numbers separated by a minus. SimplyUse the square root of each and a “+” and a “–” to get: (2x + 5)(2x – 5)

Fully factorise these:(a) 81x2 – 1(b) ¼ – t2 (c) 16y2 + 64

Answers:(a) (9x + 1)(9x – 1)(b) (½ + t)(½ – t)(c) 16(y2 + 4)

Difference of two squares

Solving quadratic equations (using factorisation)Solve this equation:

x2 + 5x – 14 = 0

(x + 7)(x – 2) = 0

x + 7 = 0 or x – 2 = 0

Factorise first

Now make each bracket equal to zero separately

2 solutionsx = - 7 or x = 2

Solve these: 2x2 + 5x - 3 =0 x2 - 7x + 10 =0 x2 + 12x + 35

=0 25t2 – 20t + 4

=0 x2 + x - 6 =0 4x2 - 64 =0

(x + 3)(2x – 1)=0(x – 5)(x – 2)=0 (x + 7)(x + 5)=0(5t – 2)(5t – 2)=0(x + 3)(x – 2)=0(2x – 8)(2x + 8)=0

x = -3 or x =1/2 x = 5 or x = 2 x = -7 or x = -5 t = 2/5 x = -3 or x = 2 x = 4 or x= -4

Solving quadratic equations (using the formula)The generalization of a quadratic equations is: ax2 + bx + c = 0The following formula works out both solutions to any quadratic equation:

x = -b ± b2 – 4ac2a

Solve 6x2 + 17x + 12 = 0 using the quadratic formula

a = 6, b = 17, c = 12

x = -b ± b2 – 4ac2a

x = -17 ± 172 – 4x6x12

2x6x = -17 ± 289 – 288

12

orx = -17 + 1

12

x = -17 - 1 12

x = -1.33.. or x = -1.5

Now solve these:1. 3x2 + 5x + 1 =02. x2 - x - 10 =03. 2x2 + x - 8 =04. 5x2 + 2x - 1 =05. 7x2 + 12x + 2 =06. 5x2 – 10x + 1 =0

(1) -0.23, -1.43 (2) 3.7, -2.7 (3) 1.77, -2.27 (4) 0.29, -0.69 (5) –0.19, -1.53 (6) 1.89, 0.11

Answers:

Solving quadratic equations (by completing the square)

Another method for solving quadratics relies on the fact that:(x + a)2 = x2 + 2ax + a2 (e.g. (x + 7)2 = x2 + 14x + 49 )

Rearranging : x2 + 2ax = (x + a)2 – a2 (e.g. x2 + 14x = (x + 7)2 – 49)

Example Rewrite x2 + 4x – 7 in the form (x + a)2 – b . Hence solve the equation x2 + 4x – 7 = 0 (1 d.p.)

Step 1 Write the first two terms x2 + 4x as a completed square x2 + 4x = (x + 2)2 – 4

Step 2 Now incorporate the third term – 7 to both sides x2 + 4x – 7 = (x + 2)2 – 4 – 7 x2 + 4x – 7 = (x + 2)2 – 11 (1st part answered)

Step 3 When x2 + 4x – 7 = 0 then (x + 2)2 – 11 = 0 (x + 2)2 = 11

x + 2 = 11 x = 11 – 2

x = 1.3 or x = - 5.3

Rearranging formulae

a

a

V = u + at

V

V

a = V - u t

x t + u

t - u

Rearrange the following formula so that a is the subject

Now rearrange these

P = 4a + 5

1.

A = be r

2.

D = g2 + c

3.

B = e + h4.

E = u - 4vd

5.

6. Q = 4cp - st

Answers: 1. a = P – 5 42. e = Ar b

3. g = D – c

4. h = (B – e)2

5. u = d(E + 4v)

6. p = Q + st 4c

Rearranging formulaeWhen the formula has the new subject in

two places (or it appears in two places during manipulation) you will need to

factorise at some point

Now rearrange these:

ab = 3a + 7

1.

s(t – r) = 2(r – 3)

3.e = u – 1 d

4.

a = e – h e + 5

2.

Rearrange to make g the subject:

Multiply all by g

Multiply out bracket

Collect all g terms on one side of the equation and factorise

(r – t) = 6 – 2s g

g(r – t) = 6 – 2gs

gr – gt = 6 – 2gs

gr – gt + 2gs = 6

g(r – t + 2s) = 6

g = 6 r – t + 2s

r = st + 6 2 + s

a = 7 b – 3

e = – h – 5a a – 1

d = u e + 1

Algebraic fractions – Addition and subtraction

Like ordinary fractions you can only add or subtract algebraic fractions if their denominators are the same

Multiply the top and bottom of

each fraction by the same amount

Show that 3 + 4 can be written as 7x + 4 x + 1 x x(x + 1)

3x + 4(x + 1) (x + 1)x x(x + 1) 3x + 4x + 4 x(x + 1) x(x + 1) 3x + 4x + 4 x(x + 1) 7x + 4 x(x + 1)

Simplify x – 6 . x – 1 x – 4

x(x – 4) – 6(x – 1) .(x – 1)(x – 4) (x – 1)(x – 4) x2 – 4x – 6x + 6 . (x – 1)(x – 4) x2 – 10x + 6 . (x – 1)(x – 4)

Algebraic fractions – Multiplication and division

Again just use normal fractions principles

6x ÷ 4x2 x2 + 4x x2 + x

Simplify:

6x × x2 + xx2 + 4x 4x2 6x × x(x + 1)x(x + 4) 4x2 6(x + 1)4x(x + 4) 3(x + 1)2x(x + 4)

23

Algebraic fractions – solving equations

4 + 7 = 2x – 2 x + 1

Solve: Multiply all by (x – 2)(x + 1)

4(x + 1) + 7(x – 2) = 2(x – 2)(x + 1)

4x + 4 + 7x – 14 = 2(x2 – 2x + x – 2)

11x – 10 = 2x2 – 4x + 2x – 4

0 = 2x2 – 13x + 6

2x2 – x – 12x + 6 = 0x(2x – 1) – 6(2x – 1) = 0(2x – 1)(x –6) = 02x – 1 = 0 or x – 6 = 0x = ½ or x = 6

Factorise

Curved graphs

y

x

y

x

y

x

y = x2

Any curve starting with x2 is “U” shaped

y = x2 3 y = x3

y = x3 + 2

Any curve starting with x3 is this shape

y = 1/x

y = 5/x

Any curve with a number /x is this shape

There are four specific types of curved graphs that you may be asked to recognise and draw.

y

x

All circles have an equation like this 16 = radius2

If you are asked to draw an accurate curved graph (eg y = x2 + 3x - 1) simply substitute x values to find y values and the co-ordinates

x2 + y2 = 16

Graphs of y = mx + c

In the equation:

y = mx + c

m = the gradient(how far up for every one along)

c = the intercept(where the line crosses the y axis)

y

x

Y = 3x + 4

13

4m

c

y

x

Graphs of y = mx + c Write down the equations of these lines:

Answers:y = xy = x + 2y = - x + 1y = - 2x + 2y = 3x + 1x = 4y = - 3

y

x

x = -2

x - 2

y = x

y < x

y = 3y > 3

Graphing inequalities

Find the region that

is not covered by these 3 regionsx - 2y xy > 3

Graphing simultaneous equations Solve these simultaneous equations using a graphical method : 2y + 6x = 12

y = 2x + 1Finding co-ordinates for 2y + 6x = 12using the “cover up” method:y = 0 2y + 6x = 12 x = 2 (2, 0)x = 0 2y + 6x = 12 y = 6 (0, 6)

Finding co-ordinates for y = 2x + 1x = 0 y = (2x0) + 1 y = 1 (0, 1)x = 1 y = (2x1) + 1 y = 3 (1, 3)x = 2 y = (2x2) + 1 y = 5 (2, 5)

y

x2

4

6

8

-8

-6

-4

1 2 3-4 -3 -2 -1 4-2

2y + 6x = 12 y = 2x + 1

The co-ordinate of the point where the two graphs cross is (1, 3).Therefore, the solutions to the simultaneous equations are:

x = 1 and y = 3

Graphical solutions to equationsIf an equation equals 0 then its solutions lie at the points where the graph of the equation crosses the x-axis.e.g. Solve the following equation graphically:

x2 + x – 6 = 0

All you do is plot the equation y = x2 + x – 6 and find where it crosses the x-axis (the line y=0)

y

x2-3

y = x2 + x – 6

There are two solutions tox2 + x – 6 = 0 x = - 3 and x =2

Graphical solutions to equationsIf the equation does not equal zero :Draw the graphs for both sides of the equation andwhere they cross is where the solutions lie

e.g. Solve the following equation graphically:x2 – 2x – 11 = 9 – x

Plot the following equations and find where they cross:y = x2 – 2x – 20 y = 9 – x

There are 2 solutions to x2 – 2x – 11 = 9 – xx = - 4 and x = 5

y

x

y = x2 – 2x – 11

y = 9 – x

-4 5

Be prepared to solve 2 simultaneous equations graphically where one is

linear (e.g. x + y = 7) and the other is a circle (e.g. x2 + y2 = 25)

Expressing laws in symbolic form

In the equation y = mx + c , if y is plotted against x the gradient of the line is m and the intercept on the y-axis is c.

Similarly in the equation y = mx2 + c , if y is plotted against x2 the gradient of the line is m and the intercept on the y-axis is c.

And in the equation y = m + c , if y is plotted against 1 the x x

gradient of the line is m and the intercept on the y-axis is c. y

xc

my = mx + cy

x2c

my = mx2 + cy

1x

cm

y = m + c x

Expressing laws in symbolic form

e.g. y and x are known to be connected by the equation y = a + b . Find a and b if: x

y 15 9 7 6 5

x 1 2 3 4 6Find the 1/x values

Plot y against 1/x1/x 1 0.5 0.33 0.25 0.17y

1/x0

5

10

15

0 0.2 0.4 0.6 0.8 1.0 1.2

xxx

x

x

b = 3

3

0.25

a = 3 ÷ 0.25 = 12 So the equation is: y = 12 + 3 x

The graph of y = - f(x) is the reflection of the graph y = f(x) in the x- axis

y

x

y = - f(x)

y = f(x)

Transformation of graphs – Rule 1

The graph of y = f(-x) is the reflection of the graph y = f(x) in the y - axis

y

x

y = f(-x)

y = f(x)

Transformation of graphs – Rule 2

y

x

y = f(x) + ay = f(x)

The graph of y = f(x) + a is the translation of the graph y = f(x) vertically by vectoro

a[ ]

Transformation of graphs – Rule 3

y

xy = f(x + a)

y = f(x)

The graph of y = f(x + a) is the translation of the graph y = f(x) horizontally by vector-a o[ ]

- a

Transformation of graphs – Rule 4

y

x

y = kf(x)

y = f(x)

The graph of y = kf(x) is the stretching of the graph y = f(x) vertically by a factor of k

Transformation of graphs – Rule 5

y

xy = f(kx)

y = f(x)

The graph of y = f(kx) is the stretching of the graph y = f(x) horizontally by a factor of 1/k

Transformation of graphs – Rule 6

Straight Distance/Time graphsD

T• The gradient of each section is the average speed for that part of the journey• The horizontal section means the vehicle has stopped

V = 0

• The section with the negative gradient shows the return journey and it will have a positive speed but a negative velocity

• Remember the rule S = D/T

V

T

Straight Velocity/Time graphs

• The gradient of each section is the acceleration for that part of the journey• The horizontal section means the vehicle is travelling at a constant velocity• The sections with a negative gradient show a deceleration

• Remember the rule A = V/T

• The area under the graph is the distance travelled